Please I want someone to answer it clearly and step by step A 30 kg bag of mail
ID: 1560783 • Letter: P
Question
Please I want someone to answer it clearly and step by step
A 30 kg bag of mail is placed at the top of a 10-m long chute inclined at 30 degree above the horizontal. The coefficient of static friction is 0.20 and the coefficient of kinetic friction is 0.15. The bag is initially at rest at the top of the chute. (a) What additional force, if any, directed along the chute is needed to hold the bag at rest? (b) If the bag is released from rest and allowed to slide down the chute, how much work is done by friction as the bag slides the length of the chute? (c) How much work is done by gravity as the bag slides the length of the chute? (d) How far does it slide along a horizontal surface at the bottom of the chute assuming the same coefficients of friction given above?Explanation / Answer
(a)
maximum static frictional force fs up the incline = us*m*g*costheta
fs = 0.2*30*9.8*cos30 = 51 N
component of gravitational force down the incline Fg = m*g*sintheta
Fg = 30*9.8*sin30 = 147 N
Fg > fs
so additonal force is required
F + f - Fg = 0
F = Fg - fs = 147 - 51 = 96 N
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(b)
kinetic friction fk = -uk*m*g
work done by friction Wf = -uk*m*g*costheta*L
Wf = -0.15*30*9.8*cos30*10 = -382 J
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(c)
work done by gravity Wg = Fg*L = m*g*sintheta*L
Wg = 30*9.8*sin30*10 = 1470 N
(d)
work done by friction along horizontal surface = uk*m*g*x = Wg - Wf
0.15*30*9.8*x = 1470 - 382
x = 24.67 m <<<<<<<-----------ANSWER